1 solutions
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0
C :
int main(int argc, char* argv[]) { double m,n,sum,i; while(~scanf("%lf%lf",&m,&n)) { sum=m; for(i=0;i<n;i++) { sum+=m; m*=1.0/2; } sum-=2*m; printf("%.2lf %.2lf\n",m,sum); } return 0; }C++ :
#include<iostream> #include<cstdio> using namespace std; int main() { double n,sum; int m; cin>>n>>m; sum=-n; for (int i=1; i<=m; i++) { sum+=2*n; n/=2; } printf("%.2lf %.2lf\n",n,sum); return 0; }
- 1
Information
- ID
- 19285
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- (None)
- # Submissions
- 0
- Accepted
- 0
- Uploaded By